HDU 1569 方格取数(2) 最大点权独立集

题目：

http://acm.hdu.edu.cn/showproblem.php?pid=1569

题意：

Problem Description

给你一个m*n的格子的棋盘，每个格子里面有一个非负数。

从中取出若干个数，使得任意的两个数所在的格子没有公共边，就是说所取数所在的2个格子不能相邻，并且取出的数的和最大。

Input

包括多个测试实例，每个测试实例包括2整数m,n和m*n个非负数(m<=50,n<=50)

Output

对于每个测试实例，输出可能取得的最大的和

思路：

hdu 1565的升级版，数据范围变大了一点，还有不是n*n的方格了，而是n*m的方格，代码稍微改一下救过了

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <set>
#include <cmath>
using namespace std;

typedef long long ll;
const int N = 3010, INF = 0x3f3f3f3f;
struct edge
{
int to, cap, next;
}g[N*300];
int cnt, head
;
int gap
, que
, level
, pre
, cur
;
int ss, tt, nv;
void add_edge(int v, int u, int cap)
{
g[cnt].to = u, g[cnt].cap = cap, g[cnt].next = head[v], head[v] = cnt++;
g[cnt].to = v, g[cnt].cap = 0, g[cnt].next = head[u], head[u] = cnt++;
}
void bfs(int t)
{
memset(level, -1, sizeof level);
memset(gap, 0, sizeof gap);
int st = 0, en = 0;
level[t] = 0;
que[en++] = t;
gap[level[t]]++;
while(st < en)
{
int v = que[st++];
for(int i = head[v]; i != -1; i = g[i].next)
{
int u = g[i].to;
if(level[u] < 0)
{
level[u] = level[v] + 1;
gap[level[u]]++;
que[en++] = u;
}
}
}
}
int sap(int s, int t)
{
bfs(t);
memcpy(cur, head, sizeof head);
int v = pre[s] = s, flow = 0, aug = INF;
while(level[s] < nv)
{
bool flag = false;
for(int &i = cur[v]; i != -1; i = g[i].next)
{
int u = g[i].to;
if(g[i].cap > 0 && level[v] == level[u] + 1)
{
flag = true;
pre[u] = v;
v = u;
aug = min(aug, g[i].cap);
if(v == t)
{
flow += aug;
while(v != s)
{
v = pre[v];
g[cur[v]].cap -= aug;
g[cur[v]^1].cap += aug;
}
aug = INF;
}
break;
}
}
if(flag) continue;
int minlevel = nv;
for(int i = head[v]; i != -1; i = g[i].next)
{
int u = g[i].to;
if(g[i].cap > 0 && level[u] < minlevel)
minlevel = level[u], cur[v] = i;
}
if(--gap[level[v]] == 0) break;
level[v] = minlevel + 1;
gap[level[v]]++;
v = pre[v];
}
return flow;
}
int main()
{
int n, m;
while(~ scanf("%d%d", &n, &m))
{
cnt = 0;
memset(head, -1, sizeof head);
int arr[60][60];
ss = 0, tt = n * m + 1;
int sum = 0;
for(int i = 1; i <= n; i++)
for(int j = 1; j <= m; j++)
{
scanf("%d", &arr[i][j]);
sum += arr[i][j];
if((i + j) % 2 == 0)
{
add_edge(ss, (i-1)*m + j, arr[i][j]);
if(i != 1) add_edge((i-1)*m + j, (i-2)*m + j, INF);
if(i != n) add_edge((i-1)*m + j, i*m + j, INF);
if(j != 1) add_edge((i-1)*m + j, (i-1)*m + j - 1, INF);
if(j != m) add_edge((i-1)*m + j, (i-1)*m + j + 1, INF);
}
else add_edge((i-1)*m + j, tt, arr[i][j]);
}
nv = tt + 1;
printf("%d\n", sum - sap(ss, tt));
}
return 0;
}